Question

please tell this answer ​

Answer 1

Answer:

Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH

and DF of angles A, B, C and D respectively and the points E, F, G

and H form a quadrilateral EFGH.

To prove that EFGH is a cyclic quadrilateral.

∠HEF = ∠AEB [Vertically opposite angles] -------- (1)

Consider triangle AEB,

∠AEB +

2

1

∠A +

2

1

∠ B = 180°

∠AEB = 180° –

2

1

(∠A + ∠ B) -------- (2)

From (1) and (2),

∠HEF = 180° –

2

1

(∠A + ∠ B) --------- (3)

Similarly, ∠HGF = 180° –

2

1

(∠C + ∠ D) -------- (4)

From 3 and 4,

∠HEF + ∠HGF = 360° –

2

1

(∠A + ∠B + ∠C + ∠ D)

= 360° –

2

1

(360°)

= 360° – 180°

= 180°

So, EFGH is a cyclic quadrilateral since the sum of the opposite

angles of the quadrilateral is 180°.]

Answer 2

Answer:

Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH

and DF of angles A, B, C and D respectively and the points E, F, G

and H form a quadrilateral EFGH.

To prove that EFGH is a cyclic quadrilateral.

∠HEF = ∠AEB [Vertically opposite angles] -------- (1)

Consider triangle AEB,

Angle AEB +1/2 angle a + 1/2 angle angle b = 180 degree

Angle AEB = 180 degree ----- 1/2 ( angle ∠A +∠ B )

From (1) and (2),

∠ HEF = 180 degree ----- 1/2 ( angle A + B )

Similarly ,  ∠ HGF = 180 degree ----- 1/2 ( angle C and angle D  )

From 3 and 4 ,

∠ HEF + ∠ HGF = 360 degree ----- 1/2 ( ∠A + ∠ B ∠ C + ∠ D )

= 360 degree ------ 1/2( 360 degree )

= 360 degree ------- 180 degree

= 180 degree

So, EFGH is a cyclic quadrilateral since the sum of the opposite

angles of the quadrilateral is 180°.

Pls mark me as brainliest

​

​

 

​

2

1

​

∠ B = 180°